bjr résoudre ces équation du deuxième degré merci d avance

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4x²-12x+9 =0

(2x-3)² =0

2x-3 =0

2x =3

x = 2/3

solution (2/3)

4x²-25x =0

(2x-5)(2x+5) =0

2x-5 =0                          2x+5 =0

2x = 5                                2x = -5

x = 5/2                                 x = -5/2

solution ( -5/2.5/2)

3xau cube = 9x²

3xauc cube -9x² =0

3x²(x-3) = 0

3x² = 0                   x-3 =0

x² =0/3                    x = 3

x² = 0

solution (0 ; 3)

e dernier je ne vois pas comment faire